Understanding Vertical Circular Motion
When an object of mass $m$ moves in a vertical circle of radius $r$, the forces acting on it (tension $T$ and weight $mg$) change as it moves around the circle. The resultant force at any point provides the necessary centripetal force ($F_c = \frac{mv^2}{r}$).
The Physics of the String
- At the top: Gravity acts downward, pulling in the same direction as the tension. Thus, the tension is at its minimum: $T_{min} = \frac{mv^2}{r} - mg$.
- At the bottom: Tension must overcome both gravity and provide the centripetal force. Thus, the tension is at its maximum: $T_{max} = \frac{mv^2}{r} + mg$.
Step-by-Step Problem Solution
Given Data:
- Mass ($m$) = 0.2 kg
- Radius ($r$) = 1 m
- Minimum Tension ($T_{min}$) = 3 N
- Assume acceleration due to gravity ($g$) = 9.8 m/s²
1. Calculate the Magnitude of Speed ($v$)
Using the formula for minimum tension at the top of the circle: $T_{min} = \frac{mv^2}{r} - mg$
Rearranging to solve for $v^2$: $\frac{mv^2}{r} = T_{min} + mg$ $v^2 = \frac{r(T_{min} + mg)}{m}$
Substitute the values: $v^2 = \frac{1(3 + (0.2 \times 9.8))}{0.2}$ $v^2 = \frac{3 + 1.96}{0.2} = \frac{4.96}{0.2} = 24.8$ $v = \sqrt{24.8} \approx 4.98 \text{ m/s}$
2. Calculate the Maximum Tension ($T_{max}$)
Using the formula for tension at the bottom of the circle: $T_{max} = \frac{mv^2}{r} + mg$
Since we know $\frac{mv^2}{r} = T_{min} + mg$, we can substitute this: $T_{max} = (T_{min} + mg) + mg = T_{min} + 2mg$
Substitute the values: $T_{max} = 3 + 2(0.2 \times 9.8)$ $T_{max} = 3 + 2(1.96) = 3 + 3.92$ $T_{max} = 6.92 \text{ N}$
Summary of Results
- Magnitude of Speed: $\approx 4.98 \text{ m/s}$
- Maximum Tension: $6.92 \text{ N}$